I have a question in deriving "time dilation" using the length contraction. Can anyone help me to find where the problem is?

Suppose the time is measured using a light traveling along a stick which is moving with a speed v with respect of the ground. The time interval in the reference frame on the stick is: t0 = L0/c, where L0 is the length of the stick measured in the reference frame of the stick, and c is the speed of the light.

Now, a person on the ground sees the stick with with length L, where L = v L0. Then the person on the ground sees the time interval for the light to travel along the stick is, t = L/c. Therefore, t = v t0. This is different from the Einstein's equation, t = t0/v. However, I couldn't find the logic problem in the above derivation. Anyone can explain it?

I have a question in deriving "time dilation" using the length contraction. Can anyone help me to find where the problem is?

Suppose the time is measured using a light traveling along a stick which is moving with a speed v with respect of the ground. The time interval in the reference frame on the stick is: t0 = L0/c, where L0 is the length of the stick measured in the reference frame of the stick, and c is the speed of the light.

Now, a person on the ground sees the stick with with length L, where L = v L0. Then the person on the ground sees the time interval for the light to travel along the stick is, t = L/c. Therefore, t = v t0. This is different from the Einstein's equation, t = t0/v. However, I couldn't find the logic problem in the above derivation. Anyone can explain it?

Sorry, I didn't specify "v" here. It is not the speed/velocity, it is the transformation coefficient, which is square root of (1-v^2/c^2). For some reason, I couldn't use all the symbols provided in my computer.

I have a question in deriving "time dilation" using the length contraction. Can anyone help me to find where the problem is?

Suppose the time is measured using a light traveling along a stick which is moving with a speed v with respect of the ground. The time interval in the reference frame on the stick is: t0 = L0/c, where L0 is the length of the stick measured in the reference frame of the stick, and c is the speed of the light.

Now, a person on the ground sees the stick with with length L, where L = v L0. Then the person on the ground sees the time interval for the light to travel along the stick is, t = L/c. Therefore, t = v t0. This is different from the Einstein's equation, t = t0/v. However, I couldn't find the logic problem in the above derivation. Anyone can explain it?

Thanks in advance!

John

If

Or the Lorentz factor,
Then for the ground observer, the length of the stick is

The person on the ground does not see the light travel along the stick in a time of L/c. The stick is moving at v, and the light moves relative to the observer at c, thus for the ground observer, the light takes t= L/(c-v) to travel the length of the stick if traveling in the same direction as the stick and L/(c+v) if traveling in the opposite direction.

This means that according to the ground observer, light traveling in one direction relative to the stick would take less time to travel the length of the stick than it does in the other. This is known as the Relativity of Simultaneity.

So let's say that you have a light source at the mid point of the Stick. It sends a light signal to two clocks that are sitting at the ends of the stick waiting for signal to start running. If you were an observer riding along with the stick the light would hit both ends at the same time and both clocks would start simultaneously. However, for our ground observer, the light signal would hit one clock before it hit the other and thus that clock will start running first, and from that point on read ahead of the other clock.
The point is that that not only will the stick and ground not agree on the length of the stick or how fast clocks run, but whether or not two events are simultaneous or not.

Thus if the light leaves the back end of the stick when a clock there reads zero, and reaches the other end when a clock there reads L0/c Then both the stick and ground observer will agree on this.
The stick says this because the clocks at both ends read the same time and this is how long it took the light to travel the length of the stick.
According to the ground this happens because even though it took more time for the light to travel the length of the stick, the clock at the leading end of the stick both read a later time when the light left the other end, but also ticked slower while the light traveled the length of the stick.

Now, if you want to derive time dilation from this, you really need to consider a clock at just one end of the stick. You can do this by putting a mirror at the other end and timing the round trip.

Now, the round trip time for the stick is 2L0/c for the stick

For the ground it is

If you take these two expressions and compare their ratios while taking into account that